Part (a)

To find the mean and standard deviation of the given z-scores:

$z3 \approx -1.101$ $z4 \approx -0.944$ $z9 \approx -0.157$ $z14 \approx 0.629$ $z20 \approx 1.573$

Mean of z-scores

The mean ($\mu_z$) of the z-scores is given by the average of the z-scores:

$\mu_z = \frac{z3 + z4 + z9 + z14 + z20}{5}$

Substitute the given values:

$\mu_z = \frac{-1.101 - 0.944 - 0.157 + 0.629 + 1.573}{5}$

$\mu_z = \frac{-1.101 - 0.944 - 0.157 + 0.629 + 1.573}{5}$

$\mu_z = \frac{-1.573 + 2.202}{5}$

$\mu_z = \frac{0.629}{5}$

$\mu_z = 0.1258$

$\mu_z \approx 0 \, (\text{rounded to the nearest whole number})$

Standard Deviation of z-scores

The standard deviation ($\sigma_z$) of the z-scores is given by:

$\sigma_z = \sqrt{\frac{(z3 - \mu_z)^2 + (z4 - \mu_z)^2 + (z9 - \mu_z)^2 + (z14 - \mu_z)^2 + (z20 - \mu_z)^2}{5}}$

Since $\mu_z \approx 0$:

$\sigma_z = \sqrt{\frac{(-1.101)^2 + (-0.944)^2 + (-0.157)^2 + (0.629)^2 + (1.573)^2}{5}}$

$\sigma_z = \sqrt{\frac{1.212 + 0.891 + 0.025 + 0.396 + 2.474}{5}}$

$\sigma_z = \sqrt{\frac{4.998}{5}}$

$\sigma_z = \sqrt{0.9996}$

$\sigma_z \approx 1 \, (\text{rounded to the nearest whole number})$

Part (b)

To find the mean and standard deviation of the given z-scores:

$z2 \approx -1.459$ $z6 \approx -0.973$ $z12 \approx -0.243$ $z17 \approx 0.365$ $z22 \approx 0.973$ $z25 \approx 1.337$

Mean of z-scores

The mean ($\mu_z$) of the z-scores is given by the average of the z-scores:

$\mu_z = \frac{z2 + z6 + z12 + z17 + z22 + z25}{6}$

Substitute the given values:

$\mu_z = \frac{-1.459 - 0.973 - 0.243 + 0.365 + 0.973 + 1.337}{6}$

$\mu_z = \frac{-2.675 + 2.675}{6}$

$\mu_z = \frac{0}{6}$

$\mu_z = 0$

Standard Deviation of z-scores

The standard deviation ($\sigma_z$) of the z-scores is given by:

$\sigma_z = \sqrt{\frac{(z2 - \mu_z)^2 + (z6 - \mu_z)^2 + (z12 - \mu_z)^2 + (z17 - \mu_z)^2 + (z22 - \mu_z)^2 + (z25 - \mu_z)^2}{6}}$

Since $\mu_z \approx 0$:

$\sigma_z = \sqrt{\frac{(-1.459)^2 + (-0.973)^2 + (-0.243)^2 + (0.365)^2 + (0.973)^2 + (1.337)^2}{6}}$

$\sigma_z = \sqrt{\frac{2.129 + 0.947 + 0.059 + 0.133 + 0.947 + 1.788}{6}}$

$\sigma_z = \sqrt{\frac{6.003}{6}}$

$\sigma_z = \sqrt{1.0005}$

$\sigma_z \approx 1 \, (\text{rounded to the nearest whole number})$

Summary

Part (a):

• Mean: 0
• Standard Deviation: 1

Part (b):

• Mean: 0
• Standard Deviation: 1

Do you want details or have any questions?

Here are 5 numbered questions you can ask:

1. How do you interpret a z-score in statistics?
2. Can you explain how to calculate the mean and standard deviation?
3. What is the significance of the standard deviation in a data set?
4. How do you find the probability corresponding to a z-score?
5. What are the properties of the normal distribution?

Tip: The mean of a set of z-scores for a standard normal distribution is always 0, and the standard deviation is always 1.